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CENTER OF MASS AND MOMENT OF INERTIA     143
determine x only.   Taking a strip of width dx for the element of mass we have
dm  cr 2 y dx
= 2 cr V2 px dx,
where cr is the mass per unit area.   Therefore substituting this expression of dm in equation (I) and changing the limits of integration we obtain
2<j f xVZpxdx
-            JQ
X =
rx$
C
JQ
2. Find the center of mass of the lamina bounded by the curves ?/2 = 4 ax and y  bx, Fig. 75. Let dx dy be the area of the element of mass, then
dm  a dx dy.
Therefore substituting in equation (I) and introducing the proper nmits of integration we obtain
/62   s*2'Vrax                                                         /&2   rz'
_    Jo   Jbx*dydx    .               _    J0   Jbx
_      0
y =
#
/62   /2Vox                                                        /62
Jo     J&*    ^d:C                                         Jo
- 6x) dx                   f62 (2 VoS - bx) dx
8a                                           2a,
:562'                                           6 *